Problem: Let $f(x)=xe^{-2x}$. What is the absolute maximum value of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{e}$ (Choice B) B $\dfrac{1}{2e}$ (Choice C) C $\dfrac{1}{e^2}$ (Choice D) D $f$ has no maximum value
Explanation: Let's first find the relative extremum points of $f$, and then consider them along with the function's end behavior in both directions. We start with finding the critical points of $f$. The derivative of $f$ is $f'(x)=e^{-2x}(1-2x)$. $f'(x)=0$ for $x=\dfrac{1}{2}$. $f'$ is defined for all real numbers. Therefore, our only critical point is $x=\dfrac{1}{2}$. Our critical point divides the function's domain (which is all real numbers) into two intervals: $\llap{-}6$ $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $x<\frac{1}{2}$ $x>\frac{1}{2}$ $\frac{1}{2}$ Let's evaluate $f'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f'(x)$ Verdict $x<\dfrac{1}{2}$ $x=0$ $f'(0)=1>0$ $f$ is increasing $\nearrow$ $x>\dfrac{1}{2}$ $x=1$ $f'(1)=-\dfrac{1}{e^2}<0$ $f$ is decreasing $\searrow$ Let's imagine ourselves walking on the graph of $f$, starting all the way to the left (from $-\infty$ ) and going all the way to the right (until $+\infty$ ). According to the table, we will start by going up and up until we reach $x=\dfrac{1}{2}$. Then, we will be forever going down. Therefore, $f$ must obtain its absolute maximum value at $x=\dfrac{1}{2}$. We are asked to find that maximum value, which is $f\left(\dfrac{1}{2}\right)=\dfrac{1}{2e}$. In conclusion, the absolute maximum value of $f$ is $\dfrac{1}{2e}$.